Co-Digestion of Swine Manure with On-Farm Organic Wastes: 22-Agric-B11 Principles of Waste Management, Undated paper, Q2

Question 2: Co-Digestion of Swine Manure with On-Farm Organic Wastes (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

1) Digester type and justification (5 marks). A mesophilic, heated, completely-mixed (CSTR) anaerobic digester is the appropriate choice. The combined feed (manure slurry plus the pretreated organic-waste stream) is diluted to roughly 3.4% total solids overall — well within the pumpable, low-solids range a conventional wet CSTR is designed for — and continuous mechanical or gas mixing is essential here specifically because the feed is heterogeneous (manure plus straw/roughage/food-waste co-substrate), which would otherwise stratify, form a floating fibrous scum layer, and settle grit at the tank bottom, all of which reduce the working volume and starve part of the microbial population of fresh substrate. Complete mixing also homogenizes temperature and pH throughout the tank, which matters more here than in a single-substrate digester because co-substrates with different buffering capacity and degradation rates can otherwise create localized pockets of volatile-fatty-acid accumulation. Operating at 35°C (mesophilic) is consistent with the source's own stated design temperature and gives a good balance of process stability and digestion rate versus the higher energy input a thermophilic (55°C) system would require.

Given.

Quantity	Value
Herd size	5,000 farrow-to-finish pigs
Dry-solid production (manure)	1.0 kg TS/pig/day
Manure slurry solids content	3% (wt)
On-farm organic-waste production	2,000 kg/day (as fed to the digester, after pretreatment)
Organic-waste total solids (after pretreatment/thermal hydrolysis)	40% (wt)
Organic-waste volatile solids	80% of TS
Digester operating temperature	35 °C (mesophilic)

Check: not supplied by the exam and fixed here as stated design assumptions — slurry/organic-waste density ≈ 1,000 kg/m³ (both streams are close to water density at these dilute-to-moderate solids contents); manure volatile solids = 80% of TS (typical swine-manure VS/TS ratio, MWPS-18); hydraulic retention time (HRT) = 20 days (typical mesophilic complete-mix range 15–20 d); two identical digesters, working depth 6 m; VS destruction = 45% (typical mesophilic baseline — thermal hydrolysis of the co-substrate typically raises this by roughly 20–30% in practice, so the boxed figures below are a conservative baseline, not an upper bound); biogas yield = 0.75 m³/kg VS destroyed.

Find. Digester volume and dimensions; daily biogas production.

Co-digestion process flow: manure slurry and pretreated organic waste feed two parallel mesophilic complete-mix digesters; recovered biogas fuels an engine-generator, with waste heat recycled to hold the digesters at 35 °C.

Approach. Convert each feed stream's dry-solids loading and solids content into a wet volumetric flow, size the combined digester volume from an assumed hydraulic retention time, then estimate daily biogas production from the combined volatile-solids loading and an assumed VS-destruction/biogas-yield pair.

Manure slurry flow. Dry solids: $\dot m_{TS,manure} = 5{,}000\ \text{pigs}\times 1.0\ \text{kg/pig}\cdot\text{d} = 5{,}000\ \text{kg TS/d}$. At 3% solids content, $$Q_{manure} = \frac{5{,}000/0.03}{1{,}000\ \text{kg/m}^3} = \boxed{166.7\ \text{m}^3/\text{d}}$$
Organic-waste flow and solids. The 2,000 kg/d organic-waste stream is 40% TS, so $TS_{organic}=2{,}000\times0.40=800\ \text{kg TS/d}$, of which 80% is volatile: $VS_{organic}=800\times0.80=\boxed{640\ \text{kg VS/d}}$. At the assumed 1,000 kg/m³ density, $Q_{organic}=2{,}000/1{,}000=2.0\ \text{m}^3/\text{d}$.
Combined feed and digester volume. Total flow $Q = 166.7+2.0=168.7\ \text{m}^3/\text{d}$; total TS $=5{,}000+800=5{,}800\ \text{kg/d}$; total VS $=(5{,}000\times0.80)+640=4{,}000+640=4{,}640\ \text{kg VS/d}$. At an assumed HRT of 20 d, $$V = Q\times\text{HRT}=168.7\times20=3{,}373.3\ \text{m}^3$$ Splitting into two identical digesters gives $V_{each}=1{,}686.7\ \text{m}^3$; at an assumed working depth of 6 m, $A=1{,}686.7/6=281.1\ \text{m}^2$, so $$D = \sqrt{\frac{4A}{\pi}} = \boxed{18.9\ \text{m}}$$ — two cylindrical digesters, each 18.9 m diameter × 6 m working depth.
Daily biogas production. At an assumed 45% VS destruction, $VS_{destroyed}=4{,}640\times0.45=2{,}088\ \text{kg/d}$. At an assumed biogas yield of 0.75 m³/kg VS destroyed, $$V_{biogas}=2{,}088\times0.75=\boxed{1{,}566\ \text{m}^3/\text{d}}$$

Quantity	Result
Manure slurry flow	166.7 m³/d
Organic-waste flow	2.0 m³/d
Combined digester feed	168.7 m³/d
Digester configuration	2 × 1,686.7 m³ (D = 18.9 m, h = 6 m)
Total digester volume	3,373.3 m³
Combined VS loading	4,640 kg VS/d
Daily biogas production	1,566 m³/d

3) Volume of digested biomass concentrated to 50% TS (5 marks).

Given. Digestion destroys only the volatile fraction of the solids (fixed/inert solids pass through unreacted), so the digestate's dry-solids mass is the feed TS minus the VS actually destroyed: $TS_{digestate}=5{,}800-2{,}088=3{,}712\ \text{kg/d}$, carried in a digestate wet mass of $168.7\ \text{m}^3/\text{d}\times1{,}000\ \text{kg/m}^3 - 2{,}088\ \text{kg/d}=166{,}579\ \text{kg/d}$ — i.e. the raw digestate leaving the digester is only about 2.2% TS.

Find. The volume of digestate after mechanical dewatering/concentration to 50% total solids.

Approach. Dry-solids mass is conserved through a dewatering step (only water is removed), so the concentrated cake's wet mass is simply the dry-solids mass divided by the target solids fraction.

Concentrate to 50% TS. $$m_{cake} = \frac{TS_{digestate}}{0.50} = \frac{3{,}712}{0.50} = 7{,}424\ \text{kg/d}$$ At an assumed concentrate density of 1,000 kg/m³, $$V_{cake} = \frac{7{,}424}{1{,}000} = \boxed{7.42\ \text{m}^3/\text{d}}$$

Quantity	Result
Digestate dry solids	3,712 kg/d
Raw digestate TS content	≈ 2.2%
Concentrated cake mass (50% TS)	7,424 kg/d
Concentrated cake volume	7.42 m³/d

Topic: Co-digestion CSTR sizing, biogas yield, and post-digestion solids concentration.

Key relations: $V=Q\times\text{HRT}$; $VS_{destroyed}=VS_{in}\times f_{destr}$; $V_{biogas}=VS_{destroyed}\times\text{yield}$; dry-solids-conserved dewatering $m_{cake}=TS/\text{target fraction}$.

Why this works: a co-digester's design volume is governed by the SAME hydraulic-retention-time logic as a single-substrate digester, just applied to the combined feed — the whole benefit of co-digestion is that a comparatively small, energy-dense organic-waste stream (2 m³/d, but 640 kg VS/d — nearly 14% of the total VS loading from just over 1% of the total flow) boosts biogas yield disproportionately to the volume it adds to the digester. Concentrating the dilute (≈2.2% TS) raw digestate afterward is a straightforward dry-mass-conservation calculation, since only water leaves in the dewatering step.

Common pitfall: applying the target 50% TS fraction to the FEED solids mass instead of the DIGESTATE solids mass (which is lower, because VS was destroyed and left as biogas) — using the feed TS would overstate the cake volume; and forgetting that VS destruction removes mass from the system as gas, so digestate wet mass is feed wet mass MINUS VS destroyed, not equal to feed wet mass.

Source: MWPS-18 Livestock Waste Facilities Handbook (anaerobic digester sizing, typical HRT/VS-destruction/biogas-yield ranges); Metcalf & Eddy Wastewater Engineering Ch. 8 (anaerobic co-digestion, solids dewatering mass balance).

Question 2 of 5: Co-Digestion of Swine Manure with On-Farm Organic Wastes

Question 2: Co-Digestion of Swine Manure with On-Farm Organic Wastes (25 marks)